## University Calculus: Early Transcendentals (3rd Edition)

$$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$
We know that $$\frac{3\pi}{2}=\pi+\frac{\pi}{2}$$ Therefore, $$\sin\frac{3\pi}{2}=\sin\Big(\pi+\frac{\pi}{2}\Big)=\sin\pi\cos\frac{\pi}{2}+\cos\pi\sin\frac{\pi}{2}=0\times0+(-1)\times1=-1$$ $$\cos\frac{3\pi}{2}=\cos\Big(\pi+\frac{\pi}{2}\Big)=\cos\pi\cos\frac{\pi}{2}-\sin\pi\sin\frac{\pi}{2}=(-1)\times0+0\times1=0$$ Now return to the given quantity and apply the addition formula for sine: $$\sin\Big(\frac{3\pi}{2}-x\Big)=\sin\frac{3\pi}{2}\cos x-\cos\frac{3\pi}{2}\sin x$$ $$\sin\Big(\frac{3\pi}{2}-x\Big)=(-1)\times\cos x-0\times\sin x$$ $$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$